Solving Fuzzy Dice Problem for Acceleration of Car

[physicsrules]

This is a standard fuzzy dice problem, hung in a car, at angle, theta, and I have to find the acceleration of the car.

I understand the approach used by the prof where he sets the x-y plane relative to mg.
So the sum of forces in the x direction = ma = Tsin(theta)
Similarly, in the y direction, Tcos(theta) = mg; so a = gtan(theta).

However, when I was going over my notes, I was wondering why can't the plane be set with respect to T and split the mg into its components. Therefore T = mgcos(theta) and ma = mgsing(theta). and obtaining a = gsin(theta). Why can't this approach be used to solve for the acceleration of the car.

 

[Doc Al]

Seems like a perfectly OK approach to me.

 

[physicsrules]

thanks for the quick reply but the acceleration is different when i use the two different methods though:
the prof's a : a = gtan(theta)
and my later pondering: a = gsin(theta)

 

[Doc Al]

but the acceleration is different when i use the two different methods though:
the prof's a : a = gtan(theta)
and my later pondering: a = gsin(theta)

Ah... you made an error. (I must not have read it all that carefully. Sorry about that!)

However, when I was going over my notes, I was wondering why can't the plane be set with respect to T and split the mg into its components. Therefore T = mgcos(theta) and ma = mgsing(theta). and obtaining a = gsin(theta). Why can't this approach be used to solve for the acceleration of the car.

You can use any axis you want. Don't forget that with the tilted axes, the acceleration will also have components:
mg\sin\theta = ma\cos\theta

Thus a = g \tan\theta for both methods.

And in the other direction (parallel to the string) you have:
T - mg\cos\theta = ma\sin\theta

 

[physicsrules]

thanks for the fast reply again! however, how is that for incline plane problems, the axes are frequently tilted, yet the acceleration does not have any component attached to it. for example, in a frictionless incline of angle theta, a = gsin(theta). how come do we not see acos(theta) = gsin(theta) for incline problems as well. Thanks!

 

[deviantdevil]

maybe it's because your a is already a component of g? so discounting normal force, there is only one force -Fg- acting on the object on an incline. in your other problem, a is a different force than g.

if I'm wrong, please do correct me

 

[Doc Al]

thanks for the fast reply again! however, how is that for incline plane problems, the axes are frequently tilted, yet the acceleration does not have any component attached to it. for example, in a frictionless incline of angle theta, a = gsin(theta). how come do we not see acos(theta) = gsin(theta) for incline problems as well.

In an incline plane problem, it's much easier to analyze if you have your axis parallel to the plane because the acceleration is parallel to the plane. (In the fuzzy dice problem, the acceleration was horizontal--that's why you needed components if you used a tilted axis.)

If you use a horizontal axis to analyze the incline plane, then you'd need to take components of acceleration and normal force.

 

[physicsrules]

ok i see it now. so in the incline problem, we are only concerned with the x- component as that's what we take into account for finding velocity and such. Then I'm assuming that the vertical component of acceleration, that we discount, merely brings the block down. Also, in the fuzzy dice problem, we have T - mgcos(theta) = masin(theta). What is the purpose of the a here? What does it do? Is it to lift the dice up, while the x-component is the acceleration that actually moves it horizontally? Thanks for clarifying this!

 

[Doc Al]

ok i see it now. so in the incline problem, we are only concerned with the x- component as that's what we take into account for finding velocity and such.

We are concerned with the component parallel to the plane (the x-component) because the block is constrained to move in that direction.

Then I'm assuming that the vertical component of acceleration, that we discount, merely brings the block down.

We're not discounting anything. The acceleration is fully in the x-direction, so that's all we need. Only if we choose to analyze the problem using vertical and horizontal axes do we have to worry about those components of the acceleration.

Also, in the fuzzy dice problem, we have T - mgcos(theta) = masin(theta). What is the purpose of the a here? What does it do? Is it to lift the dice up, while the x-component is the acceleration that actually moves it horizontally?

The acceleration comes in because we are applying Newton's 2nd law. The sum of the forces in the direction parallel to the string = T - mgcos(theta); we must set that equal to the mass times the component of the acceleration in that direction = masin(theta).

 

[physicsrules]

Thanks for your help, Doc Al.